This must be valid for any arbitrary V1, V2, ..., Vn. In particular, for V1 ≠ 0,

V2 = V3 =… = Vn = 0. Next, for V2 ≠ 0,

V1 = V3 = ... = 0, and so on. Therefore,

We conclude that

∑nj=1 yj1 = ∑nj=1 yj2 =...= ∑nj=1 yjn = 0

That is, the sum of the admittances of every column in Yi is zero.

Property 2.

The sum of the elements of every row of an indefinite admittance matrix is zero

Proof

Suppose that

V1 = V2 =...= Vn = V0 ≠

Since all the nodes are at the same potential, no currents will flow in the terminals, that is

0 = y11V0 + y12V0 +...+ y1nV0

0 = y21V0 + y22V0 +...+ y2nV0

:

0 = yn1V0 + yn2V0 +...+ ynnV0

Since V0 ≠ 0, we get

That is, the sum of the admittances of every row in Yi is zero.

Property 3

For a reciprocal n-terminal network, Yi is symmetric

yJk = ykj

Property 4

If we take the arbitrary reference (datum) as one of the terminals, say nth, then

Vn = 0

The nth column in Eq. (16.27.1) disappears and we have n equations with (n - 1) variables.

Since

In = ∑n–1k=1 Ik

and so the rath equations is redundant.

Therefore, we have (n -1) equations with (n - 1) variables, with nth node as the reference node. But this is precisely the node analysis.

Hence we conclude that if one of the nodes of an n-terminal network is made the reference node, the row and the column corresponding to the reference node are removed from the indefinite admittance matrix Yi. In this way, the short-circuit admittance is obtained.

Transtutors is the best place to get answers to all your doubts regarding properties of indefinite admittance matrix. You can submit your school, college or university level homework or assignment on network system to us and we will make sure that you get the answers you need which are timely and also cost effective.